Advanced Mathematics (Seventh Edition) Tongji University exercises 1-9 personal solutions

Advanced mathematics （ The seventh edition ） Tongji University exercises 1-9

 

$\begin{array}{rlr}& 1.seekLetterCountf(x)=\frac{x^3+3x^2-x-3}{x^2+x-6}OfevenTo\; continueDistrictbetween,andseekextremelylimit\underset{x\to 0}{\lim }f(x),\underset{x\to -3}{\lim }f(x)And\underset{x\to 2}{\lim }f(x)& \end{array}$

Explain ：

$\begin{array}{rlr}& Whenx=-3,x=2when,f(x)nothingIt\; meansThe\; righteous,theWiththistwoindividualspotbybetweenbreakspot,exceptthisAndOutside,LetterCountallevenTo\; continue,& \\ & & \\ & evenTo\; continueDistrictbetweenby(-\infty ,-3),(-3,2),(2,+\infty )& \\ & & \\ & f(x)=\frac{x^3+3x^2-x-3}{x^2+x-6}=\frac{(x+3)(x^2-1)}{(x+3)(x-2)}=\frac{x^2-1}{x-2}& \\ & & \\ & \underset{x\to 0}{\lim }f(x)=\underset{x\to 0}{\lim }\frac{x^2-1}{x-2}=\frac{1}{2}& \\ & & \\ & \underset{x\to -3}{\lim }f(x)=\underset{x\to -3}{\lim }\frac{x^2-1}{x-2}=-\frac{8}{5}& \\ & & \\ & \underset{x\to 2}{\lim }f(x)=\underset{x\to 2}{\lim }\frac{x^2-1}{x-2},and\underset{x\to 2}{\lim }\frac{x-2}{x^2-1}=0,theWith\underset{x\to 2}{\lim }f(x)=\infty & \end{array}$

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$\begin{array}{rlr}& 2.set\; upLetterCountf(x)Andg(x)stayspot{x}_{0}evenTo\; continue,ProvebrightLetterCount\phi (x)=max\{f(x),g(x)\},\psi (x)=min\{f(x),g(x)\}& \\ & & \\ & stayspot{x}_{0}alsoevenTo\; continue.& \end{array}$

Explain ：

$\begin{array}{rlr}& \phi (x)=max\{f(x),g(x)\}=\frac{1}{2}[f(x)+g(x)+|f(x)-g(x)|],& \\ & & \\ & \psi (x)=min\{f(x),g(x)\}=\frac{1}{2}[f(x)+g(x)-|f(x)-g(x)|].& \\ & & \\ & becausef(x)stayspot{x}_{0}evenTo\; continue,be|f(x)|stayspot{x}_{0}alsoevenTo\; continue;fromOnevenTo\; continueLetterCountOfand、BadstillevenTo\; continue,theWith\phi (x)、\psi (x)stayspot{x}_{0}alsoevenTo\; continue& \end{array}$

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$\begin{array}{rlr}& 3.seekNextColumnextremelylimit：& \end{array}$

$\begin{array}{rlr}& (1)\underset{x\to 0}{\lim }\sqrt{x^2-2x+5};(2)\underset{\alpha \to \frac{\pi }{4}}{\lim }(sin2\alpha {)}^3;& \\ & & \\ & (3)\underset{x\to \frac{\pi }{6}}{\lim }ln(2cos2x);(4)\underset{x\to 0}{\lim }\frac{\sqrt{x+1}-1}{x};& \\ & & \\ & (5)\underset{x\to 1}{\lim }\frac{\sqrt{5x-4}-\sqrt{x}}{x-1};(6)\underset{x\to \alpha }{\lim }\frac{sinx-sin\alpha }{x-\alpha };& \\ & & \\ & (7)\underset{x\to +\infty }{\lim }(\sqrt{x^2+x}-\sqrt{x^2-x});(8)\underset{x\to 0}{\lim }\frac{{\left(1-\frac{1}{2}{x}^2\right)}^{\frac{2}{3}}-1}{xln(1+x)}& \end{array}$

Explain ：

$\begin{array}{rlr}& (1)\underset{x\to 0}{\lim }\sqrt{x^2-2x+5}=\sqrt{\underset{x\to 0}{\lim }(x^2-2x+5)}=\sqrt{5}& \\ & & \\ & (2)\underset{\alpha \to \frac{\pi }{4}}{\lim }(sin2\alpha {)}^3={\left(\underset{\alpha \to \frac{\pi }{4}}{\lim }sin2\alpha \right)}^3={\left(sin\frac{\pi }{2}\right)}^3=1& \\ & & \\ & (3)\underset{x\to \frac{\pi }{6}}{\lim }ln(2cos2x)=ln\left(\underset{x\to \frac{\pi }{6}}{\lim }2cos2x\right)=ln\left(2cos\frac{\pi }{3}\right)=ln1=0& \\ & & \\ & (4)\underset{x\to 0}{\lim }\frac{\sqrt{x+1}-1}{x}=\underset{x\to 0}{\lim }\frac{\sqrt{x+1}-1}{(\sqrt{x+1}+1)(\sqrt{x+1}-1)}=\underset{x\to 0}{\lim }\frac{1}{\sqrt{x+1}+1}=\frac{1}{2}& \\ & & \\ & (5)\underset{x\to 1}{\lim }\frac{\sqrt{5x-4}-\sqrt{x}}{x-1}=\underset{x\to 1}{\lim }\frac{(\sqrt{5x-4}-\sqrt{x})(\sqrt{5x-4}+\sqrt{x})}{(x-1)(\sqrt{5x-4}+\sqrt{x})}=\underset{x\to 1}{\lim }\left(-\frac{x^2-5x+4}{(x-1)(\sqrt{5x-4}+\sqrt{x})}\right)=& \\ & & \\ & \underset{x\to 1}{\lim }\left(-\frac{x-4}{\sqrt{5x-4}+\sqrt{x}}\right)=\frac{3}{2}& \\ & & \\ & (6)\underset{x\to \alpha }{\lim }\frac{sinx-sin\alpha }{x-\alpha }=\underset{x\to \alpha }{\lim }\frac{2sin\frac{x-\alpha }{2}cos\frac{x+\alpha }{2}}{x-\alpha }=\underset{x\to \alpha }{\lim }\frac{sin\frac{x-\alpha }{2}}{\frac{x-\alpha }{2}}\cdot \underset{x\to \alpha }{\lim }cos\frac{x+\alpha }{2}=cos\alpha & \\ & & \\ & (7)\underset{x\to +\infty }{\lim }(\sqrt{x^2+x}-\sqrt{x^2-x})=\underset{x\to +\infty }{\lim }\frac{2x}{\sqrt{x^2+x}+\sqrt{x^2-x}}=\underset{x\to +\infty }{\lim }\frac{2}{\sqrt{1+\frac{1}{x}}+\sqrt{1-\frac{1}{x}}}=1& \\ & & \\ & (8)\underset{x\to 0}{\lim }\frac{{\left(1-\frac{1}{2}{x}^2\right)}^{\frac{2}{3}}-1}{xln(1+x)}=\underset{x\to 0}{\lim }\frac{\frac{2}{3}\cdot \left(-\frac{1}{2}{x}^2\right)}{x\cdot x}=-\frac{1}{3}& \end{array}$

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$\begin{array}{rlr}& 4.seekNextColumnextremelylimit：& \end{array}$

$\begin{array}{rlr}& (1)\underset{x\to \infty }{\lim }{e}^{\frac{1}{x}};(2)\underset{x\to 0}{\lim }ln\frac{sinx}{x};& \\ & & \\ & (3)\underset{x\to \infty }{\lim }{\left(1+\frac{1}{x}\right)}^{\frac{x}{2}};(4)\underset{x\to 0}{\lim }(1+3ta{n}^{2}x{)}^{co{t}^{2}x};& \\ & & \\ & (5)\underset{x\to \infty }{\lim }{\left(\frac{3+x}{6+x}\right)}^{\frac{x-1}{2}};(6)\underset{x\to 0}{\lim }\frac{\sqrt{1+tanx}-\sqrt{1+sinx}}{x\sqrt{1+si{n}^{2}x}-x};& \\ & & \\ & (7)\underset{x\to e}{\lim }\frac{lnx-1}{x-e};(8)\underset{x\to 0}{\lim }\frac{e^{3x}-e^{2x}-e^x+1}{\sqrt[3]{(1-x)(1+x)}-1}& \end{array}$

Explain ：

$\begin{array}{rlr}& (1)\underset{x\to \infty }{\lim }{e}^{\frac{1}{x}}=e^{\underset{x\to \infty }{\lim }\frac{1}{x}}=e^0=1& \\ & & \\ & (2)\underset{x\to 0}{\lim }ln\frac{sinx}{x}=ln\left(\underset{x\to 0}{\lim }\frac{sinx}{x}\right)=ln1=0& \\ & & \\ & (3)\underset{x\to \infty }{\lim }{\left(1+\frac{1}{x}\right)}^{\frac{x}{2}}=\underset{x\to \infty }{\lim }{\left[{\left(1+\frac{1}{x}\right)}^x\right]}^{\frac{1}{2}}=e^{\frac{1}{2}}=\sqrt{e}& \\ & & \\ & (4)\underset{x\to 0}{\lim }(1+3ta{n}^{2}x{)}^{co{t}^{2}x}=\underset{x\to 0}{\lim }[(1+3ta{n}^{2}x{)}^{\frac{1}{3}co{t}^{2}x}{]}^3=e^3& \\ & & \\ & (5)\underset{x\to \infty }{\lim }{\left(\frac{3+x}{6+x}\right)}^{\frac{x-1}{2}}=\underset{x\to \infty }{\lim }{\left[{\left(1-\frac{3}{6+x}\right)}^{-\frac{6+x}{3}}\right]}^{-\frac{3}{2}}\cdot \underset{x\to \infty }{\lim }{\left(1-\frac{3}{6+x}\right)}^{-\frac{7}{2}}=e^{-\frac{3}{2}}& \\ & & \\ & (6)\underset{x\to 0}{\lim }\frac{\sqrt{1+tanx}-\sqrt{1+sinx}}{x\sqrt{1+si{n}^{2}x}-x}=\underset{x\to 0}{\lim }\frac{tanx-sinx}{x(\sqrt{1+si{n}^{2}x}-1)(\sqrt{1+tanx}+\sqrt{1+sinx})}=& \\ & & \\ & \underset{x\to 0}{\lim }\left(\frac{sinx}{x}\cdot \frac{secx-1}{\sqrt{1+si{n}^{2}x}-1}\cdot \frac{1}{\sqrt{1+tanx}+\sqrt{1+sinx}}\right)=& \\ & & \\ & \underset{x\to 0}{\lim }\frac{sinx}{x}\cdot \underset{x\to 0}{\lim }\frac{\frac{1}{2}{x}^2}{\frac{1}{2}si{n}^{2}x}\cdot \underset{x\to 0}{\lim }\frac{1}{\sqrt{1+tanx}+\sqrt{1+sinx}}=\frac{1}{2}& \\ & & \\ & (7)Maket=x-e,bex=t+e,Whenx\to ewhen,t\to 0,\underset{x\to e}{\lim }\frac{lnx-1}{x-e}=\underset{t\to 0}{\lim }\frac{ln(t+e)-lne}{t}=\underset{t\to 0}{\lim }\frac{ln\left(1+\frac{t}{e}\right)}{t}=\frac{1}{e}& \\ & & \\ & (8)\underset{x\to 0}{\lim }\frac{e^{3x}-e^{2x}-e^x+1}{\sqrt[3]{(1-x)(1+x)}-1}=\underset{x\to 0}{\lim }\frac{(e^{2x}-1)(e^x-1)}{(1-x^2{)}^{\frac{1}{3}}-1}=\underset{x\to 0}{\lim }\frac{2x\cdot x}{-\frac{1}{3}{x}^2}=-6& \end{array}$

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$\begin{array}{rlr}& 5.set\; upf(x)stayROnevenTo\; continue,Andf(x)\ne 0,\phi (x)stayROnYessetThe\; righteous,Andfrombetweenbreakspot,beNextColumnChenStatementinwhichsomeyesYesOf,& \\ & & \\ & whichsomeyeswrongOf？Such\; asfruityesYesOf,trysaybrightThe\; reason\; isfrom;Such\; asfruityeswrongOf,trytoOutwanttobackexample.& \end{array}$

$\begin{array}{rlr}& (1)\phi [f(x)]have\; toYesbetweenbreakspot;(2)[\phi (x){]}^{2}have\; toYesbetweenbreakspot;& \\ & & \\ & (3)f[\phi (x)]nothave\; toYesbetweenbreakspot;(4)\frac{\phi (x)}{f(x)}have\; toYesbetweenbreakspot& \end{array}$

Explain ：

$\begin{array}{rlr}& (1)wrongOf,\phi (x)=sgnx,f(x)=e^x,\phi [f(x)]\equiv 1stayROnevenTo\; continue& \\ & & \\ & (2)wrongOf,\phi (x)=\{\begin{array}{l}1,x\in Q,\\ \\ -1,x\in R\text{}Q,\end{array}[\phi (x){]}^2\equiv 1stayROnevenTo\; continue& \\ & & \\ & (3)YesOf,\phi (x)=\{\begin{array}{l}1,x\in Q,\\ \\ -1,x\in R\text{}Q,\end{array}f(x)=|x|+1,f[\phi (x)]\equiv 2stayROnevenTo\; continue& \\ & & \\ & (4)YesOf,Such\; asfruitF(x)=\frac{\phi (x)}{f(x)}stayROnevenTo\; continue,be\phi (x)=F(x)\cdot f(x)alsostayROnevenTo\; continue,AndhasknowSpearshield& \end{array}$

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$\begin{array}{rlr}& 6.set\; upLetterCountf(x)=\{\begin{array}{l}{e}^x,x<0,\\ \\ \alpha +x,x\ge 0\end{array}Should\; beWhenhowsamplechooseChooseCount\alpha ,onlycansendhave\; tof(x)becomebystay(-\infty ,+\infty )InsideOfevenTo\; continueLetterCount.& \end{array}$

Explain ：

   from On first etc. Letter Count Of even To continue sex , f ( x ) stay ( − ∞ , 0 ) and ( 0 , + ∞ ) Inside even To continue , the With want send f ( x ) stay ( − ∞ , + ∞ ) Inside even To continue ,    only want choose Choose Count α , send f ( x ) stay x = 0 It's about even To continue namely can .    stay x = 0 It's about , lim ⁡ x → 0 − f ( x ) = lim ⁡ x → 0 − e x = 1 , lim ⁡ x → 0 + f ( x ) = lim ⁡ x → 0 + ( α + x ) = α , f ( 0 ) = α ,    take α = 1 , be lim ⁡ x → 0 − f ( x ) = lim ⁡ x → 0 + f ( x ) = f ( 0 ) = 1 , the With take α = 1 , f ( x ) Just become by stay ( − ∞ , + ∞ ) Inside Of even To continue Letter Count \begin{aligned} &\ \ Because of the continuity of elementary functions ,f(x) stay (-\infty,0) and (0,+\infty) Internal continuity , So we should make f(x) stay (-\infty,+\infty) Internal continuity ,\\\\ &\ \ Just choose the number \alpha, send f(x) stay x=0 It can be continuous .\\\\ &\ \ stay x=0 It's about ,\lim_{x \rightarrow 0^-}f(x)=\lim_{x \rightarrow 0^-}e^x=1,\lim_{x \rightarrow 0^+}f(x)=\lim_{x \rightarrow 0^+}(\alpha+x)=\alpha,f(0)=\alpha,\\\\ &\ \ take \alpha=1, be \lim_{x \rightarrow 0^-}f(x)=\lim_{x \rightarrow 0^+}f(x)=f(0)=1, So take \alpha=1,f(x) Become in (-\infty,+\infty) Continuous functions in & \end{aligned} ​   from On first etc. Letter Count Of even To continue sex ,f(x) stay (−∞,0) and (0,+∞) Inside even To continue , the With want send f(x) stay (−∞,+∞) Inside even To continue ,   only want choose Choose Count α, send f(x) stay x=0 It's about even To continue namely can .   stay x=0 It's about ,x→0−lim​f(x)=x→0−lim​ex=1,x→0+lim​f(x)=x→0+lim​(α+x)=α,f(0)=α,   take α=1, be x→0−lim​f(x)=x→0+lim​f(x)=f(0)=1, the With take α=1,f(x) Just become by stay (−∞,+∞) Inside Of even To continue Letter Count ​​